# integral proofing

#### cummings123321

let f(x,t)=xe^(-xt).show that the integral I(x)=∫f(x,t)dt (integration from 0 to infinite)exists for all x>=0 . is x->I(x) continuous on [0,infinite)
what should i use here to prove the integral exist ???can someone give me the detail expanlation???

#### Plato

MHF Helper
let f(x,t)=xe^(-xt).show that the integral I(x)=∫f(x,t)dt (integration from 0 to infinite)exists for all x>=0 . is x->I(x) continuous on [0,infinite) what should i use here to prove the integral exist ???can someone give me the detail expanlation???
$$\displaystyle \int_0^\infty {xe^{ - xt} dt} = \lim _{b \to \infty } \left( {\left. { - e^{ - xt} } \right|_{t = 0}^{t = b} } \right)=\lim _{b \to \infty } \left( {1 - e^{-xb} } \right) = ?$$

#### cummings123321

thank you,then the limit is 1 so the integral exists,but how about the continuous part, for now x->I(x) ,I(X)=1 it seems ,it is continuous?? i am not sure

#### Plato

MHF Helper
thank you,then the limit is 1 so the integral exists,but how about the continuous part, for now x->I(x) ,I(X)=1 it seems ,it is continuous?? i am not sure
Frankly, I have the same question about $$\displaystyle I(x)$$.
We know that $$\displaystyle I(x)\ge 0$$. But I have no idea how it fits into the question.

#### hollywood

Isn't $$\displaystyle I(x)=1$$ for $$\displaystyle x>0$$ and $$\displaystyle I(x)=0$$ for $$\displaystyle x=0$$? So it would be continuous on $$\displaystyle (0,\infty)$$ and discontinuous at 0, right?

- Hollywood